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The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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19 Jul 2017, 22:49
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The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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20 Jul 2017, 02:03
Bunuel wrote: The square root of integer x is equal to the sum of the cubes of y and z. If the squares of y and z are each less than 10, what is the greatest possible value of x?
A. 2916 B. 3981 C. 3982 D. 3999 E. 4000 Hi... Let the values be √10 each. here the values will be just LESS than 10, so our ACTUAL answer will be slightly less than the answer we get here So sum of cubes is \(2*10^{\frac{3}{2}}\) Square of this is \(4*10^3=4000\).. But the actual ans will be slightly LESS than 4000 and an integer.. So 3999 D
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The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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25 Jul 2017, 09:46
chetan2u wrote: Bunuel wrote: The square root of integer x is equal to the sum of the cubes of y and z. If the squares of y and z are each less than 10, what is the greatest possible value of x?
A. 2916 B. 3981 C. 3982 D. 3999 E. 4000 Hi... Let the values be √10 each. here the values will be just LESS than 10, so our ACTUAL answer will be slightly less than the answer we get here So sum of cubes is \(2*10^{\frac{3}{2}}\) Square of this is \(4*10^3=4000\).. But the actual ans will be slightly LESS than 4000 and an integer.. So 3999 D Chetan  Thanks for responding. Although, I wish you hadn't used "yellow" to highlight the text against an almost yellow background. Nevertheless, here is my approach and I don't know if I arrived at the wrong answer. Given, \(\sqrt{x}\) = \(y^3\) + \(z^3\) And is also given, \(y^2\) < 10 \(z^2\) < 10 So, the biggest value that y and z could each take is 3. Since, the question asks for the highest value, I have assumed the highest value for y and z (i.e, 3). \(\sqrt{x}\) = \(3^3\) + \(3^3\) => \(\sqrt{x}\) = 27 + 27 => \(\sqrt{x}\) = 54 and squaring both sides, we get: x = \({54}^2\) x = 2916. Answer = A Please correct if I am wrong.



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Re: The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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25 Jul 2017, 10:28
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Bunuel wrote: The square root of integer x is equal to the sum of the cubes of y and z. If the squares of y and z are each less than 10, what is the greatest possible value of x?
A. 2916 B. 3981 C. 3982 D. 3999 E. 4000 We can create the following equation: √x = y^3 + z^3 Since the squares of y and z are each less than 10, y < √10 and z < √10, which implies y^3 < 10√10 and z^3 < 10√10. Thus: √x < 10√10 +10√10 √x < 20√10 x < 4000 Since x is an integer, the greatest value of x is 3999. Answer: D
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Re: The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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25 Jul 2017, 11:27
JeffTargetTestPrep wrote: . . . y < √10 and z < √10, which implies y^3 < 10√10 and z^3 < 10√10 . . . Beats me why I should "root" the y instead of assuming a possible highest value for y in \(y^2\) e.g., y= 3 ?



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Re: The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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25 Jul 2017, 11:34
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Blackbox wrote: JeffTargetTestPrep wrote: . . . y < √10 and z < √10, which implies y^3 < 10√10 and z^3 < 10√10 . . . Beats me why I should "root" the y instead of assuming a possible highest value for y in \(y^2\) e.g., y= 3 ? Y and z need not be integers Sent from my SMG610F using GMAT Club Forum mobile app



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Re: The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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25 Jul 2017, 17:55
Blackbox wrote: chetan2u wrote: Bunuel wrote: The square root of integer x is equal to the sum of the cubes of y and z. If the squares of y and z are each less than 10, what is the greatest possible value of x?
A. 2916 B. 3981 C. 3982 D. 3999 E. 4000 Hi... Let the values be √10 each. here the values will be just LESS than 10, so our ACTUAL answer will be slightly less than the answer we get here So sum of cubes is \(2*10^{\frac{3}{2}}\) Square of this is \(4*10^3=4000\).. But the actual ans will be slightly LESS than 4000 and an integer.. So 3999 D Chetan  Thanks for responding. Although, I wish you hadn't used "yellow" to highlight the text against an almost yellow background. Nevertheless, here is my approach and I don't know if I arrived at the wrong answer. Given, \(\sqrt{x}\) = \(y^3\) + \(z^3\) And is also given, \(y^2\) < 10 \(z^2\) < 10 So, the biggest value that y and z could each take is 3. Since, the question asks for the highest value, I have assumed the highest value for y and z (i.e, 3). \(\sqrt{x}\) = \(3^3\) + \(3^3\) => \(\sqrt{x}\) = 27 + 27 => \(\sqrt{x}\) = 54 and squaring both sides, we get: x = \({54}^2\) x = 2916. Answer = A Please correct if I am wrong. Hi, Sorry for the colour. Didn't realise it. You are WRONG when you read it as an integer. It is nowhere given as an integer. So if √10=3.16227766.., the value could be 3.16227765.. Hence we find answer by taking it as 10 and finding one value lesser INTEGER. Hope it helps
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The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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25 Jul 2017, 18:52
perfect squares less than 10 are 1 ,4,9 for which value of y,z could be 1 ,2 ,3 but for greatest possible value of x we will take y= z= 3 now √x = y^3+z^3 =3^3 + 3^3 √x = 27+27 = 54 x= 54^2 x=2916 Answer A
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Re: The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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26 Jul 2017, 05:56
Perfect squares less than 10 are 1 ,4,9 for which value of y,z could be 1 ,2 ,3 but for greatest possible value of x we will take y= z= 3 now √x = y^3+z^3 =3^3 + 3^3 √x = 27+27 = 54 x= 54^2 x=2916 Option A.
Problem states square root of integer x is equal to the sum of the cubes of y and z, hence to achieve that we cannot have y & z to be non integer values as we frame the equation x=y^3+z^3=10√10+10√10, as √10 when simplified would give a decimal component as well. Or in other words y & z cannot be √10 each. Please advise.



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Re: The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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26 Jul 2017, 06:53
Hi FB2017Although I did the same as you did, but I was wrong. below is explanation we are not saying x=10√10+10√10 rather it is x<10√10+10√10 i.e x<20√10 (Note: if we take x=20√10 then we will get x=4000) as x<20√10 we are saying that x<4000 i.e x=3999 as x should be an integer. Kudo please if this helps you.
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The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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14 Aug 2017, 01:41
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Bunuel wrote: The square root of integer x is equal to the sum of the cubes of y and z. If the squares of y and z are each less than 10, what is the greatest possible value of x?
A. 2916 B. 3981 C. 3982 D. 3999 E. 4000 \(\sqrt{x} = y^3 + z^3\) \(y^2 <10 , z^2 <10\) \(> y <\sqrt{10}, z <\sqrt{10}\) So, \(\sqrt{x} < \sqrt{10}^3 + \sqrt{10}^3\) \(\sqrt{x} < 20*\sqrt{10}\) \(x < (20*\sqrt{10})^2\) \(x<4000\) \(So x = 3999\) Answer D
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The square root of integer x is equal to the sum of the cubes of y and [#permalink]
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14 Aug 2017, 02:23
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Ans is clearly Dconsider max values y^2=10 means y^6 =1000 and same for z^6=1000. or y=z x^0.5 = y^3+z^3 square both sides x= y^6+z^6+2(yz)^3 = 1000+1000+2(z^2)^3 = 1000=1000+2(z^6) =1000+1000+2(1000) = 4000 but y and z are less than 10 therefore x is less than 4000 and integer value less than 4000 is 3999. Give kudos take kudos
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